Integration - two sin functions

[exidez]

Homework Statement

I have the answer I just don't know how he got it! I want to know how he got from the second line to the third line in the image below.

Homework Equations

The Attempt at a Solution

I have used integration by part and the u substitution method.

When substituting with u i am still left with the other terms in the other sine function.
When integrating by parts, it just doesn't simplify..

Is there another method i don't know about? There doesn't seem to be any trig identities either.

 

[Дьявол]

Hi & welcome to PF!

Use the product to sum formulae:

Regards.

 

[exidez]

ok, that seems to be the way to go (i must remember these)
but i am still having problems evaluating it...

\frac{1}{2a^2}\int cos(at-2a\tau) d\tau - \int cos(at) d\tau

\frac{1}{2a^2}[\frac{sin(at-2a\tau)}{-2a}]from t to 0 - [cos(at)\tau] from t to 0

im going to stop there because something is wrong already...

 

[Дьявол]

Are a and t constants?

Obviously \int cos(x)dx = sin(x).

Regards.

 

[g_edgar]

\sin(at-a\tau)\sin(a\tau) = ?

 

[exidez]

sin(at-a\tau)sin(a\tau) = \frac{1}{2}(cos(at-a\tau-a\tau) - cos(at-a\tau+a\tau)) = \frac{1}{2}(cos(at-2a\tau) - cos(at))
subbing this into the integral and taking the half outside:

\frac{1}{2a^2}( \int cos(at-2a\tau) d\tau - \int cos(at) d\tau )

i treated cos(at) as a constant because we are integrating with respect to tau, not t...
or is this where i am going wrong?

 

[Дьявол]

sin(at-a\tau)sin(a\tau) = \frac{1}{2}(cos(at-a\tau-a\tau) - cos(at-a\tau+a\tau)) = \frac{1}{2}(cos(at-2a\tau) - cos(at))
subbing this into the integral and taking the half outside:

\frac{1}{2a^2}( \int cos(at-2a\tau) d\tau - \int cos(at) d\tau )

i treated cos(at) as a constant because we are integrating with respect to tau, not t...
or is this where i am going wrong?

Actually, you forgot the limits. So your integral would be:

\frac{1}{2a^2}( \int_{0}^{t} cos(at-2a\tau) d\tau - \int_{0}^{t} cos(at) d\tau)

Now make a substitution,

- for the first integral w=at-2a\tau

- for the second integral treat cos(at) as constant, so that:


\frac{1}{2a^2}( \int_{0}^{t} cos(at-2a\tau) d\tau - cos(at)\int_{0}^{t} d\tau)

Now find dw but do not forget the limits of the integral. You need to substitute 0, t for tau, in w.

Regards.

 

[exidez]

Ok here's another attempt

\frac{1}{2a^2}( \int_{0}^{t} cos(at-2a\tau) d\tau - cos(at)\int_{0}^{t} d\tau)

let u = at-2a\tau
\frac{du}{d\tau} = -2a
\frac{du}{-2a} = d\tau

\frac{1}{2a^2}( \int_{0}^{t} cos(u) \frac{du}{-2a} - cos(at)\int_{0}^{t} d\tau)\frac{1}{2a^2}(\frac{1}{-2a} \int_{0}^{t} cos(u) du - cos(at)\int_{0}^{t} d\tau)

\frac{1}{2a^2}(\frac{1}{-2a} \int_{0}^{t} cos(u) du - cos(at)(t-0))

\frac{1}{2a^2}(\frac{1}{-2a} \int_{0}^{t} cos(u) du - tcos(at))

substituting u back in and evaluating

\frac{1}{2a^2}(\frac{1}{-2a} (sin(at-2at) - sin(at)) - tcos(at))\frac{1}{2a^2}(\frac{sin(at-2at)}{-2a} - \frac{sin(at)}{-2a} - tcos(at))\frac{1}{2a^3}(\frac{sin(at-2at)}{-2} - \frac{sin(at)}{-2} - atcos(at))

-\frac{1}{4a^3}(sin(at-2at) - sin(at)} + 2atcos(at))im stuck again, somehow i don't think that 4 is suppose to be in front there and the coefficient of 2 in front of the cos shouldn't be there either... is there another trig identity i should know?

 

[Дьявол]

As I said, again you missed the limit.

\frac{1}{2a^2}( \int_{0}^{t} cos(at-2a\tau) d\tau - cos(at)\int_{0}^{t} d\tau)

let u = at-2a\tau
\frac{du}{d\tau} = -2a
\frac{du}{-2a} = d\tau

Now, for \tau=0, u=at and for \tau=t, u = at-2at=-at

So your integral should look like: <br /> \frac{1}{2a^2}(\frac{1}{-2a} \int_{at}^{-at} cos(u) du - cos(at)\int_{0}^{t} d\tau) <br />

Be careful with those limits!

Regards.

 

[exidez]

Oh, i now realize what you ment!
i always forget about those when i substitute u!
thanks will try again

 

[exidez]

That help! Thankyou! You don't know how long i have been trying to figure this out.

you posted the limits the wrong way round, but you probably mixed them up when you typed it into the brackets {}{}..


\frac{1}{2a^2}(\frac{1}{-2a} \int_{at}^{-at} cos(u) du - cos(at)\int_{0}^{t} d\tau)

\frac{1}{2a^2}(\frac{1}{-2a} \int_{at}^{-at} cos(u) du - cos(at)(t-0))

\frac{1}{2a^2}(\frac{1}{-2a} \int_{at}^{-at} cos(u) du - tcos(at))

\frac{1}{2a^2}(\frac{1}{-2a} (sin(-at) - sin(at)) - tcos(at))

\frac{1}{2a^2}(\frac{1}{-2a} (- sin(at) - sin(at)) - tcos(at))

\frac{1}{2a^2}(\frac{1}{a} (sin(at) - tcos(at))

\frac{1}{2a^3}(sin(at) - atcos(at)

Thanks again. I can now sleep knowing how he does it now! ;)

 

[Дьявол]

Yep, you did it right.

Regards.